Question

pls prove that--
sin(A+B)sin(A-B)+sin(B+C)sin(B-C)+sin(C+A)-sin(C-A)=0

Answer 1

myininaya will do this

Answer 2

sin(a+b)sin(a-b):
\[(\sin(a)\cos(b)+\sin(b)\cos(a))(\sin(a)\cos(b)-\sin(b)\cos(a)) \]
\[= \sin^2(a)\cos^2(b)-\sin^2(b)\cos^2(a)\]
each one of the terms will become a difference of two squares. i dont know if this leads to a solution or not, seems like a decent start though...gotta read >.<

Answer 3

thank u joe
let me go once more with it

Answer 4

will u pls say the steps

Answer 5

ok
thank u myininaya

Answer 6

thats blank

Answer 7

lol

Answer 8

yeah........!!!!!!

Answer 9

ha...ha...

Answer 10

thanks a lot myininaya
i ll go through it

Answer 11

hey i made a mistake in that attachment

Answer 12

bt i could not reach the answer
i ll post another qstion
pls anwser

Answer 13

i;m still working on it

Answer 14

so i have this so far
\[\cos^2B(\sin^2A-\sin^2C)-\sin^2B(\cos^2A-\cos^2C)+2cosCsinA\]

Answer 15

so this is s trig class?

Answer 16

yeah.........

Answer 17

i hav a doubt that whether i had any mistake in writing the qstion
ie, instead of sin(A+B)sin(A-B)+sin(B+C)sin(B-C)+sin(C+A)-sin(C-A)=0,
it is sin(A+B)sin(A-B)+sin(B+C)sin(B-C)+sin(C+A)sin(C-A)=0

Answer 18

oh no really lol

Answer 19

well then i got it lol

Answer 20

just funny
!!!!

Answer 21

the parts where i X out just pretend that part is sin(c+a)sin(c-a)

Answer 22

ya i got it.........
thank u for ur concern

Answer 23

the trick was to expand all of them and multiply then i wrote everything in terms of sin
by using identities mainly sin^2x+cos^2x=1 identity.